Let (2α, α) be the largest interval in which the function f(t) = |t+1|/t², t

Q: Why do we split the function at $t = -1$ instead of just differentiating the modulus directly?

Great question! The modulus function $|t+1|$ is not differentiable at $t = -1$. By splitting the function into two cases, we replace the modulus with a simple algebraic expression, allowing us to use the quotient rule safely and correctly.

Q: I got two values for $x$ when solving $g'(x) = 0$. Why do we reject $x = 1$?

Always keep an eye on the domain! The function $g(x)$ is defined as $2 \ln(x - 2) + \dots$ for $x > 2$. Since $x = 1$ is outside this domain, it is physically impossible for the function to exist there. We must discard it.

Q: How do we know for sure that $x = 3$ is a local maximum and not a minimum?

We use the second derivative test. By calculating $g''(x) = -\frac{2}{(x-2)^2} - 2$, we see that for any $x > 2$, the second derivative is always negative. A negative second derivative means the curve is concave down, which guarantees a local maximum.

Question

Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2 \log_e(x - 2) + \alpha x^2 + 4x - \alpha, x > 2$, is .........

Sigma Percentile · Accepted Answer

### Analyzing the Function $f(t)$ The function is defined as $f(t) = \frac{|t+1|}{t^2}$. Because the modulus sign creates a point of non-differentiability at $t = -1$, we must analyze the function in two distinct regimes: $t < -1$ and $-1 \le t < 0$. For the region $t < -1$, the expression $t+1$ is negative, which forces the modulus to reveal itself as $-(t+1)$. This transforms our function into: $$f(t) = \frac{-(t+1)}{t^2} = \frac{-t-1}{t^2}$$ ### Finding the Monotonicity Applying the quotient rule to $f(t) = \frac{-t-1}{t^2}$, we calculate the derivative: $$f'(t) = \frac{(-1)(t^2) - (-t-1)(2t)}{(t^2)^2} = \frac{-t^2 + 2t^2 + 2t}{t^4} = \frac{t^2 + 2t}{t^4} = \frac{t+2}{t^3}$$ In the territory where $t < -1$, the denominator $t^3$ is always negative. For the function to be strictly decreasing, we require $f'(t) < 0$, which implies the numerator $t+2$ must be positive. This leads us to the interval $(-2, -1)$. Comparing this to the given interval $(2\alpha, \alpha)$, we deduce that $2\alpha = -2$ and $\alpha = -1$. Thus, we find **$\alpha = -1$**. ### Analyzing the Function $g(x)$ With $\alpha$ secured, we examine the function $g(x) = 2 \ln(x - 2) - x^2 + 4x + 1$. This function is defined for $x > 2$, which serves as our strict boundary. To find the local maximum, we seek the critical points where the slope of the tangent, $g'(x)$, is zero: $$g'(x) = \frac{2}{x-2} - 2x + 4$$ Setting $g'(x) = 0$ leads to the following algebraic manipulation: $$\frac{2}{x-2} = 2x - 4$$ $$\frac{2}{x-2} = 2(x-2)$$ $$(x-2)^2 = 1$$ ### Final Calculation and Verification Solving $(x-2)^2 = 1$ yields $x-2 = 1$ or $x-2 = -1$, resulting in $x = 3$ or $x = 1$. Given our boundary constraint $x > 2$, we discard $x = 1$ and accept **$x = 3$**. To confirm this is a local maximum, we use the second derivative test: $$g''(x) = -\frac{2}{(x-2)^2} - 2$$ At $x = 3$, we find $g''(3) = -\frac{2}{1} - 2 = -4$. Since $g''(3) < 0$, the curve is concave down, confirming a local maximum. Substituting $x = 3$ back into $g(x)$: $$g(3) = 2 \ln(3-2) - (3)^2 + 4(3) + 1$$ $$g(3) = 2 \ln(1) - 9 + 12 + 1 = 0 + 4 = 4$$ The local maximum value of the function is **$4$**.

Enter Numerical Value:

Visualized Solution

Analyzing $f (t)$ for $t < 0$

Case 1: $- 1 \leq t < 0$

Differentiating for Case 1

Monotonicity in Case 1

Case 2: $t < - 1$

Differentiating for Case 2

Finding the Decreasing Interval

Determining $α$

Defining $g (x)$

Finding Critical Points of $g (x)$

Solving for $x$

Verifying Local Maximum

Calculating Local Maximum Value

The Sigma Insight: Maxima and Minima

Analyzing the Function $f (t)$

Finding the Monotonicity

Analyzing the Function $g (x)$

Final Calculation and Verification

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Enter Numerical Value:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t<0

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −1≤t<0

Differentiating for Case 1

Monotonicity in Case 1

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t<−1

Differentiating for Case 2

Finding the Decreasing Interval

Determining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Finding Critical Points of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Verifying Local Maximum

Calculating Local Maximum Value

The Sigma Insight: Maxima and Minima

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Finding the Monotonicity

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Final Calculation and Verification

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